Problem: The 2050 volunteers are organizing the "Run! Chase the Rising Sun" activity. Starting on Apr 25 at 7:30 am, runners will complete the 6km trail around the Yunqi town.

There are n+1 checkpoints on the trail. They are numbered by 0, 1, ..., n. A runner must start at checkpoint 0 and finish at checkpoint n. No checkpoint is skippable — he must run from checkpoint 0 to checkpoint 1, then from checkpoint 1 to checkpoint 2 and so on. Look at the picture in notes section for clarification.

Between any two adjacent checkpoints, there are m different paths to choose. For any 1≤ i≤ n, to run from checkpoint i-1 to checkpoint i, a runner can choose exactly one from the m possible paths. The length of the j-th path between checkpoint i-1 and i is b_{i,j} for any 1≤ j≤ m and 1≤ i≤ n.

To test the trail, we have m runners. Each runner must run from the checkpoint 0 to the checkpoint n once, visiting all the checkpoints. Every path between every pair of adjacent checkpoints needs to be ran by exactly one runner. If a runner chooses the path of length l_i between checkpoint i-1 and i (1≤ i≤ n), his tiredness is $$$min_{i=1}^n l_i,$$$ i. e. the minimum length of the paths he takes.

Please arrange the paths of the m runners to minimize the sum of tiredness of them.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10 000). Description of the test cases follows.

The first line of each test case contains two integers n and m (1 ≤ n,m ≤ 100).

The i-th of the next n lines contains m integers b_{i,1}, b_{i,2}, ..., b_{i,m} (1 ≤ b_{i,j} ≤ 10^9).

It is guaranteed that the sum of n⋅ m over all test cases does not exceed 10^4.

Output

For each test case, output n lines. The j-th number in the i-th line should contain the length of the path that runner j chooses to run from checkpoint i-1 to checkpoint i. There should be exactly m integers in the i-th line and these integers should form a permuatation of b_{i, 1}, ..., b_{i, m} for all 1≤ i≤ n.

If there are multiple answers, print any.

Example

Input


2
2 3
2 3 4
1 3 5
3 2
2 3
4 1
3 5


Output


2 3 4
5 3 1
2 3
4 1
3 5

Note

In the first case, the sum of tiredness is min(2,5) + min(3,3) + min(4,1) = 6.

<image>

In the second case, the sum of tiredness is min(2,4,3) + min(3,1,5) = 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class B {

	public static BufferedReader br;
	public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

	public static void main(String[] args) throws IOException {
		readInput();
		out.close();
	}

	public static void readInput() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		// br = new BufferedReader(new FileReader("in.in"));
		int t = Integer.parseInt(br.readLine());
		while (t-->0) {
			StringTokenizer st = new StringTokenizer(br.readLine());
			int n = Integer.parseInt(st.nextToken());
			int m = Integer.parseInt(st.nextToken());
			List<Long>[] ts = new ArrayList[n];
			for (int i = 0 ; i < n; i++) {
				st = new StringTokenizer(br.readLine());
				ts[i] = new ArrayList<Long>();
				for (int j = 0 ; j < m; j++) {
					ts[i].add(Long.parseLong(st.nextToken()));
				}
				Collections.sort(ts[i]);
			}
			long[][] ans = new long[n][m];
			for (int te =0; te < m; te++) {
				// Get the minimum then select the maximum for the rest.
				int mi = 0;
				for (int i = 0; i < n; i++) {
					if (ts[mi].get(0) > ts[i].get(0)) mi = i;
				}
				for (int i = 0; i < n; i++) {
					if (i == mi) {
						ans[i][te] = ts[i].remove(0);
					}
					else ans[i][te] = ts[i].remove(ts[i].size()-1);
				}
			}
			for (long[] a: ans) {
				for (long b: a) {
					out.print(b + " ");
				}
				out.println();
			}
		}
	}
}